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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2000

  • question_answer
    A bomb is fixed from a canon with a velocity of 1000 m/s making an angle of\[30{}^\circ \]with the horizontal\[(g=9.8\text{ }m/{{s}^{2}})\]. Time taken by bomb to reach the highest point is:

    A)  40 s                      

    B)         30 s

    C)  51 s                      

    D)         25 s

    E)  15s

    Correct Answer: C

    Solution :

    \[T=\frac{u\,\sin \theta }{g}\] Given, \[u=1000\text{ }m/s,\]                 \[\theta =30{}^\circ ,g=9.8m/{{s}^{2}}\]                 \[T=\frac{1000\times \sin 30{}^\circ }{9.8}=51\,s\]


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